Author Topic: VCE Methods Question Thread! (Read 5860870 times) Tweet Share

jessss0407 · « **Reply #6420 on:** October 31, 2014, 04:43:37 pm »

Hey guys!

could someone help me with this?

Q. Sharelle is the goal shooter for her netball team. During her matches, she has many attempts at scoring a goal. Assume that each attempt at scoring a goal is independent of any other attempt. In the long term, her scoring rate has been shown to be 80% (that is, 8 out of 10 attempts to score a goal are successful). What is the probability, correct to three decimal places, that her first 4 attempts at scoring a goal are successful, given that exactly 6 of her first 8 attempts at scoring a goal in a match are successful?

Thanks!

Professor_Oak · « **Reply #6421 on:** October 31, 2014, 04:51:25 pm »

With finding the inverse function, do we have to write the function in this way? Or can I just write the rule and then write dom h-1=... after?

Orb · « **Reply #6422 on:** October 31, 2014, 05:00:51 pm »

Quote from: jessss0407 on October 31, 2014, 04:43:37 pm

Hey guys!

could someone help me with this?

Q. Sharelle is the goal shooter for her netball team. During her matches, she has many attempts at scoring a goal. Assume that each attempt at scoring a goal is independent of any other attempt. In the long term, her scoring rate has been shown to be 80% (that is, 8 out of 10 attempts to score a goal are successful). What is the probability, correct to three decimal places, that her first 4 attempts at scoring a goal are successful, given that exactly 6 of her first 8 attempts at scoring a goal in a match are successful?

Thanks!

Let S = Success
Pr(S) = 0.8
You want to find Pr(4S) given that Pr(6 of 8 are successful).

This is also [Pr(4S) intersection Pr(6of8)]/Pr(6of8) --> The Pr(AintB)/Pr(B) formula

To find Pr(4S):
Every shot has to be successful in the first four tries, so we only calculate the probability of succeeding, that is:
(0.8 )^4 = 0.4096

To find Pr(6of8):
This is a binomial distribution. So input into your calc binomCdf(8,0.8,6,8) = 0.7969

To find Pr(4S)intPr(6of8):
This is just the probability of 4 successes coupled with 2 of 4 successes in the next 4 tries.
= (0.8 )^4 x binomPdf(4,0.8,2)
= 0.0629

Final Calculation: 0.0629/0.7969 = 0.07895

fadzsta1 · « **Reply #6423 on:** October 31, 2014, 06:15:11 pm »

Hey guys, can anyone help with this question? I know theres a change of base involved somewhere but I cant figure where to use it. Any assistance would be appreciated.

Solve the equation $log_{x}5 + log_{5}x = \frac{5}{2}$ for x.

Brunette15 · « **Reply #6424 on:** October 31, 2014, 07:37:17 pm »

Can someone please explain what the difference is between C and D? :/

keltingmeith · « **Reply #6425 on:** October 31, 2014, 07:42:22 pm »

Quote from: Brunette15 on October 31, 2014, 07:37:17 pm

Can someone please explain what the difference is between C and D? :/

Very interesting - you see, if C is true, D is true - but, if D is true, that doesn't mean that C is true - so the answer cannot be C.

Let me explain:

A function is NOT differentiable if:
-The function is not continuous
-The gradient of the limits on both sides at that point are the same (i.e, sharp points and end points)

So, if D is true (which it is), then this could be because of any of the above. Assuming C is not true and the function IS continuous there, then the derivative doesn't exist because we're at a sharp point or an end point. In this case, it's a sharp point.

Phy124 · « **Reply #6426 on:** October 31, 2014, 08:01:04 pm »

Quote from: fadzsta1 on October 31, 2014, 06:15:11 pm

Hey guys, can anyone help with this question? I know theres a change of base involved somewhere but I cant figure where to use it. Any assistance would be appreciated.

Solve the equation $log_{x}5 + log_{5}x = \frac{5}{2}$ for x.

This is one way (hopefully the long way $:-\$ ) of doing it

$\log_{5}(x) + \log_{x}(5) = \frac{5}{2}$

$\Leftrightarrow \frac{\log_{e}(x)}{\log_e(5)} + \frac{\log_{e}(5)}{\log_e(x)}} = \frac{5}{2}$

$\Leftrightarrow \frac{\log_{e}(x)\log_e(x)+\log_e(5)\log_{e}(5)}{\log_e(5)\log_e(x)}= \frac{5}{2}$

$\Leftrightarrow (\log_{e}(x))^2+(\log_e(5))^2=\frac{5}{2}\log_e(5)\log_e(x)$

$\Leftrightarrow (\log_{e}(x))^2-\frac{5}{2}\log_e(5)\log_e(x)+(\log_e(5))^2=0$

$\text{Let} \ A = \log_{e}(x)$

$A^2-\frac{5}{2}\log_e(5)A+(\log_e(5))^2=0$

$\Leftrightarrow A = \frac{\frac{5}{2}\log_e(5)\pm\sqrt{(\frac{5}{2}\log_e(5))^2-4(1)(\log_e(5))^2}}{2}$

$\Leftrightarrow A = \frac{\frac{5}{2}\log_e(5)\pm\sqrt{\frac{25}{4}(\log_e(5))^2-4(\log_e(5))^2}}{2}$

$\Leftrightarrow A = \frac{\frac{5}{2}\log_e(5)\pm\sqrt{\frac{9}{4}(\log_e(5))^2}}{2}$

$\Leftrightarrow A = \frac{\frac{5}{2}\log_e(5)\pm\frac{3}{2}\log_e(5)}{2}$

$\Leftrightarrow A = 2\log_e(5),\frac{1}{2}\log_{e}(5)$

$\Leftrightarrow A = \log_e(5^2),\log_{e}(5^{1/2})$

$\Leftrightarrow A = \log_e(25),\log_{e}(\sqrt{5})$

$\Leftrightarrow \log_{e}(x)= \log_e(25),\log_{e}(\sqrt{5})$

$\Rightarrow x = \sqrt{5},25$

*Note: in the very first step I switched the position of the two terms on the left hand side because I misread their order in latex, it doesn't change anything else, though (solution is fine)

speedy · « **Reply #6427 on:** October 31, 2014, 08:08:41 pm »

Quote from: IndefatigableLover on October 30, 2014, 11:01:59 pm

Haha just did this paper today and luckily I got this question right!

So essentially from your answer from Question 2e. you would have gotten 8/3 using -> $(\frac{32}{3}-{\frac{28}{3})+(\frac{44}{3}-{\frac{40}{3})$ however realistically Tasmania Jones has one shot to get the diamond since he can't drop it halfway and pick it up afterwards and all... so basically your answer from Question 2e is halved to 4/3 (just chose one of the fractions I guess) which is equal to 80 minutes when you convert from hours to minutes giving you the overall total time From there you would just do this:

$\therefore 80-\frac{325}{8}-19-(\frac{325}{8}\cdot\frac{1}{2})$

$=\frac{1}{16}\text{min}$

Therefore he has 1/16 minutes left to spare or 3.75 seconds when you convert from minutes to seconds

Do you understand the very last probability question? aha

Reus · « **Reply #6428 on:** October 31, 2014, 08:54:13 pm »

Best tips on what to do for last minute study? What exams are the best to do and is a bound reference important? Thanks.

keltingmeith · « **Reply #6429 on:** October 31, 2014, 09:34:58 pm »

Quote from: Reus on October 31, 2014, 08:54:13 pm

Best tips on what to do for last minute study? What exams are the best to do and is a bound reference important? Thanks.

If by last minute you mean now, make sure you know everything and crank out the most recent VCAA papers.
If by last minute you mean the night before, ONLY revise material. DO NOT do practice questions, PARTICULARLY if you don't have WORKED SOLUTIONS for the questions.

Bound reference is very nice to have! Even if you don't have something made up, at least take in your textbook. It's like a dictionary in an English exam - you probably won't use it, but if you need to and don't have it, you'll be spewin'.

Reus · « **Reply #6430 on:** October 31, 2014, 09:51:08 pm »

Quote from: EulerFan101 on October 31, 2014, 09:34:58 pm

If by last minute you mean now, make sure you know everything and crank out the most recent VCAA papers.
If by last minute you mean the night before, ONLY revise material. DO NOT do practice questions, PARTICULARLY if you don't have WORKED SOLUTIONS for the questions.

Bound reference is very nice to have! Even if you don't have something made up, at least take in your textbook. It's like a dictionary in an English exam - you probably won't use it, but if you need to and don't have it, you'll be spewin'.

Thanks! Yes I mean last minute revision as in tonight till Wednesday haha. Which VCAA exams do you recommend the most?

psyxwar · « **Reply #6431 on:** October 31, 2014, 09:52:56 pm »

Quote from: fadzsta1 on October 31, 2014, 06:15:11 pm

Hey guys, can anyone help with this question? I know theres a change of base involved somewhere but I cant figure where to use it. Any assistance would be appreciated.

Solve the equation $log_{x}5 + log_{5}x = \frac{5}{2}$ for x.

Quote from: Phy124 on October 31, 2014, 08:01:04 pm

This is one way (hopefully the long way $:-\$ ) of doing it
Spoiler
$\log_{5}(x) + \log_{x}(5) = \frac{5}{2}$

$\Leftrightarrow \frac{\log_{e}(x)}{\log_e(5)} + \frac{\log_{e}(5)}{\log_e(x)}} = \frac{5}{2}$

$\Leftrightarrow \frac{\log_{e}(x)\log_e(x)+\log_e(5)\log_{e}(5)}{\log_e(5)\log_e(x)}= \frac{5}{2}$

$\Leftrightarrow (\log_{e}(x))^2+(\log_e(5))^2=\frac{5}{2}\log_e(5)\log_e(x)$

$\Leftrightarrow (\log_{e}(x))^2-\frac{5}{2}\log_e(5)\log_e(x)+(\log_e(5))^2=0$

$\text{Let} \ A = \log_{e}(x)$

$A^2-\frac{5}{2}\log_e(5)A+(\log_e(5))^2=0$

$\Leftrightarrow A = \frac{\frac{5}{2}\log_e(5)\pm\sqrt{(\frac{5}{2}\log_e(5))^2-4(1)(\log_e(5))^2}}{2}$

$\Leftrightarrow A = \frac{\frac{5}{2}\log_e(5)\pm\sqrt{\frac{25}{4}(\log_e(5))^2-4(\log_e(5))^2}}{2}$

$\Leftrightarrow A = \frac{\frac{5}{2}\log_e(5)\pm\sqrt{\frac{9}{4}(\log_e(5))^2}}{2}$

$\Leftrightarrow A = \frac{\frac{5}{2}\log_e(5)\pm\frac{3}{2}\log_e(5)}{2}$

$\Leftrightarrow A = 2\log_e(5),\frac{1}{2}\log_{e}(5)$

$\Leftrightarrow A = \log_e(5^2),\log_{e}(5^{1/2})$

$\Leftrightarrow A = \log_e(25),\log_{e}(\sqrt{5})$

$\Leftrightarrow \log_{e}(x)= \log_e(25),\log_{e}(\sqrt{5})$

$\Rightarrow x = \sqrt{5},25$

*Note: in the very first step I switched the position of the two terms on the left hand side because I misread their order in latex, it doesn't change anything else, though (solution is fine)

$\log_{x}(5) + \log_{5}(x) = \frac{5}{2}$

$\text{Using change of base formula:}$

$\frac{1}{\log_{5}(x)} + \log_{5}(x)=\frac{5}{2}$

$\text{Let} \ A = \log_{5}(x)$

$\Leftrightarrow \frac{1}{\text{A}} + A=\frac{5}{2}$

$\Leftrightarrow 1+A^2=\frac{5}{2}A$

$\Leftrightarrow A^2-\frac{5}{2}A+1=0$

$\text{Then solve for x using quadratic formula}$

$\Rightarrow x = \sqrt{5},25$

Easier method

Blondie21 · « **Reply #6432 on:** October 31, 2014, 09:54:42 pm »

How do you get from the left equation to the right one?

Like I found the equation of the left but I couldn't solve it to equal that (even though I they are were equal)

keltingmeith · « **Reply #6433 on:** October 31, 2014, 09:55:55 pm »

Quote from: Reus on October 31, 2014, 09:51:08 pm

Thanks! Yes I mean last minute revision as in tonight till Wednesday haha. Which VCAA exams do you recommend the most?

All of the most recent ones.

Go from last year backwards. The closer they are to now, the more relevant they are to your exam.

keltingmeith · « **Reply #6434 on:** October 31, 2014, 09:57:17 pm »

Quote from: Blondie21 on October 31, 2014, 09:54:42 pm

How do you get from the left equation to the right one?

Like I found the equation of the left but I couldn't solve it to equal that (even though I they are were equal)

What they did is: $\cos^2(x)=\sqrt{(\cos^2(x))^2}=\sqrt{\cos^4(x)}$ , then multiplied that into the root. I honestly don't know why, though...

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